Simplify the following expression: $y = \dfrac{5x^2- 1x- 4}{x - 1}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(5)}{(-4)} &=& -20 \\ {a} + {b} &=& &=& {-1} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-20$ and add them together. Remember, since $-20$ is negative, one of the factors must be negative. The factors that add up to ${-1}$ will be your ${a}$ and ${b}$ When ${a}$ is ${4}$ and ${b}$ is ${-5}$ $ \begin{eqnarray} {ab} &=& ({4})({-5}) &=& -20 \\ {a} + {b} &=& {4} + {-5} &=& -1 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({5}x^2 +{4}x) + ({-5}x {-4}) $ Factor out the common factors: $ x(5x + 4) - 1(5x + 4)$ Now factor out $(5x + 4)$ $ (5x + 4)(x - 1)$ The original expression can therefore be written: $ \dfrac{(5x + 4)(x - 1)}{x - 1}$ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ This leaves us with $5x + 4; x \neq 1$.